Question

Solid sodium reacts with liquid water to form hydrogen gas according

to the equation
2 Na(s) + 2 H₂O(1)→ 2 NaOH(aq) + H₂(g)
What is the pressure of hydrogen gas in the 20.0 L headspace of a
reactor vessel when 1.00 kg sodium is reacted with excess water at
50.0°C?

Answer 1

To find the pressure of hydrogen gas produced from sodium reacting with water, the ideal gas law is applied after determining moles of hydrogen from the mass of sodium and given reaction stoichiometry.

Step-by-step explanation:

The question pertains to the chemical reaction between sodium and water to produce hydrogen gas and sodium hydroxide (NaOH).

To find the pressure of the hydrogen gas produced, we can use the ideal gas law, PV=nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

First, calculate the number of moles of sodium (Na) used. The molar mass of sodium is approximately 23 g/mol, so 1.00 kg (1000 g) of sodium is 1000 g / 23 g/mol = 43.48 mol.

According to the balanced chemical equation, 2 moles of Na produce 1 mole of H2, so the reaction will produce 43.48 / 2 = 21.74 moles of hydrogen.

Next, convert the temperature to Kelvin by adding 273.15 to the Celsius temperature: 50.0°C + 273.15 = 323.15 K.

Now, using the ideal gas law and solving for P, where R is 0.0821 L·atm/(K·mol), and V is 20.0 L:

P = (nRT)/V = (21.74 mol × 0.0821 L·atm/(K·mol) × 323.15 K) / 20.0 L = 28.6 atm (rounded to two decimal places).